Finding Limit of f(i,z) for i_0, z_0

[azizz]

Homework Statement

For the function:

$$f(i,z) := \frac{1}{z-z_0} \left( \sqrt{g} + \sqrt{\frac{k}{m}} \frac{i-i_0}{z-z_0} \right)$$

we have to find a solution for $$f(i_0,z_0)$$.

Homework Equations

$$z_0 = \sqrt{\frac{k}{mg}} i_0$$

$$i_0 = 1$$

The Attempt at a Solution

As you can see the system goes to $$\infty$$ if we let $$z$$ and $$i$$ go to $$z_0$$ and $$i_0$$ at the same time.

So my fist guess was to first let $$i$$ go to $$i_0$$, this gives

$$f(i_0,z) := \frac{1}{z-z_0} \sqrt{g}$$

But again for $$z \rightarrow z_0$$ the function $$f(i_0,z_0)$$ will go to $$\infty$$ (even if I fill in the equation for $$z_0$$ as given above; no cancellations take place).

4. Hints from the professor

I asked my professor for a couple of hints, but I am still unable to solve the problem with this information. Perhaps you can use it?

- it you let both go to its nominal value ($$z_0$$ or $$i_0$$) at the same time then the function will go to $$\infty$$. But there exists a certain trajectory (fixed $$i=i_0$$ and $$z(i) \rightarrow z_0$$ or vv) where the function has a solution.

- so you cannot let $$z$$ and $$i$$ go to $$z_0$$ and $$i_0$$ at the same time. Instead you have to let one of the two go to its nominal value ($$z_0$$ or $$i_0$$) and then let the other go to its nominal value as a function of the other.


Anyone has a suggestion how this problem can be tackled?

 

[Dick]

How about making sure that sqrt(k/m)*(i-i0)/(z-z0) is always equal to -sqrt(g)?

 

[azizz]

Thanks for the fast reply! That might be a solution indeed, because then f(i0,z0)=0. I haven't checked this yet, but in the remainder of the exercise I have to let f(i0,z0) vary 50% from its nominal value (that is f(i0,z0)). So I don't expect this to be the solutions, but I will take a look at it.

 

[Dick]

I rather suspect if you are clever about the path, you can make it converge to anything you want. That's what makes the problem not very interesting.

 

[azizz]

I can see f(i0,z0)=0 is a solution indeed. I can't see that other values are solutions as well, but I can imagine your right. However, for the remainder of the exercise I expect only one solution to be the desired one. Is it possible that there exist something like a "best" solution (I doubt this though).

 

[Dick]

Think about solving sqrt(g)+sqrt(k/m)*(i-i0)/(z-z0)=(z-z0)/K for i as a function of z and let i->i0. Wouldn't that give you a limit of K?

 

[azizz]

Im sorry, but I still don't see how this is to be done. First of all, why introduce the new variable K? Isnt that the same as 1/f(i,z)?

Then you say that the limit of K can be found by solving the given equation for i as function of z, that would be

$$i = \sqrt{\frac{m}{k}} (z-z_0) \left( \frac{z-z_0}{K} - \sqrt{g} \right) + i_0$$

So for $$z \rightarrow z_0$$ we get $$i \rightarrow i_0$$, but we knew that already...

 

[Dick]

The point is to make limit of f(i,z) as i->i0, z->z0 equal to 1/K. Where K is pretty much anything we want. Your function f just plain doesn't have a definite limit. I'm not sure I can think of any interesting questions to ask about the 'limit'.

 

[HallsofIvy]

You say "limit" in the title but then you say just "find f(i₀,z₀)" in the body of you post. That value is, of course, NOT defined. Do you mean to define f(i₀,z₀) as the limit so that the function is continuous? (And again, as you have been told, that limit does not exist so this is impossible.)

 

[azizz]

yes that is right, the limit does not exist. I misused this term. Indeed I am looking for a value for f(i0,z0) such that the function is continuous at this point.